Answer
$L(x)=-5$ at $x=x_{0}$
Work Step by Step
$f(x)=2x^2+4x-3 $,
$L(x)=f(a)+f'(a)(x-a)$
$f'(x)=4x+4$
at $x_{0}=-1$ the function and its derivative is close to the value -0.9
$x_{0}=-1$
$f(-1)=2(-1)^2+4(-1)-3=-5$
$f'(-1)=4(-1)+4=0$
then $L(x)=-5-0(x+1)=-5$
$L(x)=-5$ at $x=x_{0}$
Thus, the final answer is: $L(x)=-5$ at $x=x_{0}$
